Questions & Answers

Question

Answers

Answer
Verified

Hint: Assume two positive integers $n\ and\ \left( n+1 \right)$. Multiply them together. After multiplication, prove that the product obtained can be written in the form $2K$where $'K'$ is any positive integer by using mathematical induction.

__Complete step-by-step answer:__

According to the question, we have to prove that the product of two positive integers is divisible by 2.

Let us assume two positive integers $n\ and\ \left( n+1 \right)$.

Their product will be $n\times \left( n+1 \right)={{n}^{2}}+n$.

Now, we have to prove that ${{n}^{2}}+n$ is even. Let us use mathematical induction to prove that ${{n}^{2}}+n$ is divisible by 2.

Let,

$\begin{align}

& P\left( n \right)={{n}^{2}}+n \\

& \Rightarrow P\left( 1 \right)=1+1=2 \\

\end{align}$

$P\left( 1 \right)=2$, so $P\left( 1 \right)$ is divisible by 2 and $P\left( 1 \right)$is even.

Let us assume that $P\left( r \right)$ is even.

$i.e.\ ''P\left( r \right)={{r}^{2}}+r''$is even.

So, ${{r}^{2}}+r$ can be written as $2{{K}_{1}}$ where ${{K}_{1}}$ is any positive integer.

${{r}^{2}}+r=2{{K}_{1}}.............\left( 1 \right)$

Now, let us see $P\left( r+1 \right)$,

$\begin{align}

& P\left( r+1 \right)={{\left( r+1 \right)}^{2}}+\left( r+1 \right) \\

& ={{r}^{2}}+2r+1+r+1 \\

\end{align}$

Writing $2r\ as\ r+r$, we will get,

$\begin{align}

& P\left( r+1 \right)={{r}^{2}}+\left( r+r \right)+r+2 \\

& P\left( r+1 \right)=\left( {{r}^{2}}+r \right)+2r+2 \\

& P\left( r+1 \right)=\left( {{r}^{2}}+r \right)+2\left( r+1 \right) \\

\end{align}$

From equation (1), we can write ${{r}^{2}}+r=2{{k}_{1}}$ and we can write $\left( r+1 \right)={{k}_{2}}$.

$\begin{align}

& \Rightarrow P\left( r+1 \right)=2{{K}_{1}}+2{{K}_{2}} \\

& \Rightarrow P\left( r+1 \right)=2\left( {{K}_{1}}+{{K}_{2}} \right) \\

& \Rightarrow P\left( r+1 \right)=2\left( {{K}_{3}} \right)\ \ \ \ \ \ \left[ \text{Assuming }{{K}_{1}}+{{K}_{2}}={{K}_{3}} \right] \\

\end{align}$

As, $P\left( r+1 \right)$ can be written as 2 multiplied by a constant, so $P\left( r+1 \right)$ will be even.

$P\left( 1 \right)$ is even.

And $P\left( r+1 \right)$ is even if $P\left( r \right)$ is even.

So, by mathematical induction, it is proved that ${{n}^{2}}+n$ is even for any positive integer $n$.

Note: In two consecutive numbers, one of the numbers will be odd and another will be even. Product of an odd number and an even number will always be even.

We can write an even number as $2n$ and an odd number as $\left( 2n+1 \right)$.

Then their product,

$\begin{align}

& =\left( 2n \right)\times \left( 2n+1 \right) \\

& =2\times n\times \left( 2n+1 \right) \\

& =2K \\

\end{align}$

So, the product will be even.

According to the question, we have to prove that the product of two positive integers is divisible by 2.

Let us assume two positive integers $n\ and\ \left( n+1 \right)$.

Their product will be $n\times \left( n+1 \right)={{n}^{2}}+n$.

Now, we have to prove that ${{n}^{2}}+n$ is even. Let us use mathematical induction to prove that ${{n}^{2}}+n$ is divisible by 2.

Let,

$\begin{align}

& P\left( n \right)={{n}^{2}}+n \\

& \Rightarrow P\left( 1 \right)=1+1=2 \\

\end{align}$

$P\left( 1 \right)=2$, so $P\left( 1 \right)$ is divisible by 2 and $P\left( 1 \right)$is even.

Let us assume that $P\left( r \right)$ is even.

$i.e.\ ''P\left( r \right)={{r}^{2}}+r''$is even.

So, ${{r}^{2}}+r$ can be written as $2{{K}_{1}}$ where ${{K}_{1}}$ is any positive integer.

${{r}^{2}}+r=2{{K}_{1}}.............\left( 1 \right)$

Now, let us see $P\left( r+1 \right)$,

$\begin{align}

& P\left( r+1 \right)={{\left( r+1 \right)}^{2}}+\left( r+1 \right) \\

& ={{r}^{2}}+2r+1+r+1 \\

\end{align}$

Writing $2r\ as\ r+r$, we will get,

$\begin{align}

& P\left( r+1 \right)={{r}^{2}}+\left( r+r \right)+r+2 \\

& P\left( r+1 \right)=\left( {{r}^{2}}+r \right)+2r+2 \\

& P\left( r+1 \right)=\left( {{r}^{2}}+r \right)+2\left( r+1 \right) \\

\end{align}$

From equation (1), we can write ${{r}^{2}}+r=2{{k}_{1}}$ and we can write $\left( r+1 \right)={{k}_{2}}$.

$\begin{align}

& \Rightarrow P\left( r+1 \right)=2{{K}_{1}}+2{{K}_{2}} \\

& \Rightarrow P\left( r+1 \right)=2\left( {{K}_{1}}+{{K}_{2}} \right) \\

& \Rightarrow P\left( r+1 \right)=2\left( {{K}_{3}} \right)\ \ \ \ \ \ \left[ \text{Assuming }{{K}_{1}}+{{K}_{2}}={{K}_{3}} \right] \\

\end{align}$

As, $P\left( r+1 \right)$ can be written as 2 multiplied by a constant, so $P\left( r+1 \right)$ will be even.

$P\left( 1 \right)$ is even.

And $P\left( r+1 \right)$ is even if $P\left( r \right)$ is even.

So, by mathematical induction, it is proved that ${{n}^{2}}+n$ is even for any positive integer $n$.

Note: In two consecutive numbers, one of the numbers will be odd and another will be even. Product of an odd number and an even number will always be even.

We can write an even number as $2n$ and an odd number as $\left( 2n+1 \right)$.

Then their product,

$\begin{align}

& =\left( 2n \right)\times \left( 2n+1 \right) \\

& =2\times n\times \left( 2n+1 \right) \\

& =2K \\

\end{align}$

So, the product will be even.